For certain metal incident frequency `nu` is five times threshold frequency `nu_(0)` and the maximum velocity of the photoelectrons is `8xx10^(6)ms^(-1)`. If incident photon frequency is `2nu_(0)`, the maximum velocity of photoelectrons will be

Options

A. `4 xx 10^(6) ms^(-1)`
B. `6 xx 10^(6) ms^(-1)`
C. `8 xx 10^(6) ms^(-1)`
D. `1 xx 10^(6) ms^(-1)`

Answer

A

Explanation

According to Einstein.s photoelectric equation
`hupsilon=hupsilon_0+1/2mv_(max)^2" or " 1/2mv_(max)^2=hupsilon-hupsilon_0`
According to the given problem
`1/2 m(8xx10^(6))^2 = (5 upsilon-upsilon_0) " "..(i)`
`1/2 mv_(max)^2 =h (2upsilon_0 -upsilon_0) " "..(ii)`
Dividing equation (i) by (ii)
`v_(max)^2 =((8xx10^6)^2)/4 " or " v_(max) = 4 xx10^(6) ms^(-1)`